Assume now that \(G\)  is irreducible and primitive. Let \(D(G)\)  be the derived subgroup. By [1]}, either \(D(G)\)  is \(\mathrm {SL}_3(C)\) , either it is isomorphic to \(\mathrm {PSL}_2(C)\) . Since \(\mathrm {SO}_3(C)\)  is also a primitive subgroup of \(\mathrm {SL}_3(C)\) , it is isomorphic to \(\mathrm {PSL}_2(C)\) .
The matrices commuting with the two mentioned groups are homotecies and we find that the center is isomorphic to an algebraic subgroup of \(C^*\) . Then the center of \(G\)  is isomorphic to \(\det (G)^{1/3}=\lbrace  c\in C^*| c^3 \in \det (G)\rbrace  \) , with the writing
\(A\mapsto \det (A)^{1/3}\times A\det (A)^{-1/3}\)  (note that we may take the roots of elements of \(C\)  since the latter is algebraically closed).
So let us write
\(G\sim \det (G)^{1/3} D(G)\) .
The group \(\det (G)\)  may be computed with an equation of order one, \(\phi (y)=\det (A)y\) .
We want to give a criterion to decide whether \(D(G)=\mathrm {SO}_3(C)\)  or not. Let us prove the following lemma.
